Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+3y &= 1 \\ 3x-2y &= 1\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $3x = 2y+1$ Divide both sides by $3$ to isolate $x$ $x = {\dfrac{2}{3}y + \dfrac{1}{3}}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{2}{3}y + \dfrac{1}{3}}) + 3y = 1$ $-\dfrac{16}{3}y - \dfrac{8}{3} + 3y = 1$ Simplify by combining terms, then solve for $y$ $-\dfrac{7}{3}y - \dfrac{8}{3} = 1$ $-\dfrac{7}{3}y = \dfrac{11}{3}$ $y = -\dfrac{11}{7}$ Substitute $-\dfrac{11}{7}$ for $y$ in the top equation. $-8x+3( -\dfrac{11}{7}) = 1$ $-8x-\dfrac{33}{7} = 1$ $-8x = \dfrac{40}{7}$ $x = -\dfrac{5}{7}$ The solution is $\enspace x = -\dfrac{5}{7}, \enspace y = -\dfrac{11}{7}$.